A stationary $ \mathrm{He}^{+} $ion emitted a photon corresponding to first line of lyman series. The electron liberated from stationary H- atom after this elastic collision will have certain de Broglie wave length $ (\lambda) . \lambda $ will be expressed as $ (0.26 \times \mathrm{x}) Å $ for photo electron Take $ \sqrt{5.51}=2.34 $. The value of $ \mathrm{x} $ will be

Question

A stationary $ \mathrm{He}^{+} $ion emitted a photon corresponding to first line of lyman series. The electron liberated from stationary H- atom after this elastic collision will have certain de Broglie wave length $ (\lambda) . \lambda $ will be expressed as $ (0.26 \times \mathrm{x}) Å $ for photo electron Take $ \sqrt{5.51}=2.34 $. The value of $ \mathrm{x} $ will be,

MARKS App · Accepted Answer

Photon emitted from He + = - 13.6 2 2 1 2 2 - 1 = 40.8 eV Liberated electron energy = 40.8 - 13.6 ∵ IE of H atom = 13.6 eV i.e. as if accelerated by electric potential of 27.2 volts . λ in Å for photo electron = 1 5 0 27.2 = 5.51 = 2.34 2.34 = 0.26 × x ∴ x = 9

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